在为每个执行 a 时从 std::vector 中擦除?

本文介绍了在为每个执行 a 时从 std::vector 中擦除?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着跟版网的小编来一起学习吧！

问题描述

限时送ChatGPT账号..

迭代的正确方法是使用迭代器.但是，我认为通过擦除，迭代器无效.

The proper way to iterate is to use iterators. However, I think by erasing, the iterator is invalidated.

基本上我想做的是:

for(iterator it = begin; it != end; ++it)
{
    if(it->somecondition() )
    {
     erase it
    }

}

如果没有 v[i] 方法，我怎么能做到这一点?

How could I do this without v[i] method?

谢谢

struct RemoveTimedEvent
{
    bool operator()(const AguiTimedEvent& pX, AguiWidgetBase* widget) const 
    {
        return pX.getCaller() == widget;
    }
};

void AguiWidgetContainer::clearTimedEvents( AguiWidgetBase* widget )
{
    std::vector<AguiTimedEvent>::iterator it = std::remove_if(timedEvents.begin(),
        timedEvents.end(), RemoveTimedEvent());
    timedEvents.erase(it, timedEvents.end());

}

推荐答案

erase() 返回一个新的迭代器:

erase() returns a new iterator:

for(iterator it = begin; it != end(container) /* !!! */;)
{
    if (it->somecondition())
    {
        it = vec.erase(it);  // Returns the new iterator to continue from.
    }
    else
    {
        ++it;
    }
}

请注意，我们不能再将它与预先计算的结束进行比较，因为我们可能会删除它并因此使其无效.我们每次都必须明确地结束.

Note that we can no longer compare it against a precalculated end, because we may erase it and therefore invalidate it. We must get the end explicitly each time.

更好的方法可能是结合 std::remove_if 和 erase().你从 O(N²)(每个元素都会被擦除并随着你的移动而移动)变成 O(N):

A better method might be to combine std::remove_if and erase(). You change from being O(N²) (every element gets erased and shifted as you go) to O(N):

iterator it = std::remove_if(begin, end, pred);
vec.erase(it, vec.end());

其中 pred 是您的移除谓词，例如:

Where pred is your removal predicate, such as:

struct predicate // do choose a better name
{
    bool operator()(const T& pX) const // replace T with your type
    {
        return pX.shouldIBeRemoved();
    }
};

iterator it = std::remove_if(begin, end, predicate());
vec.erase(it, vec.end());

在你的情况下，你可以让它变得非常通用:

In your case, you can make it pretty general:

class remove_by_caller
{
public:
    remove_by_caller(AguiWidgetBase* pWidget) :
    mWidget(pWidget)
    {}

    // if every thing that has getCaller has a base, use that instead
    template <typename T> // for now a template
    bool operator()(const T& pX) const
    {
        return pX.getCaller() == mWidget;
    }

private:
    AguiWidgetBase* mWidget;
};

std::vector<AguiTimedEvent>::iterator it =
    std::remove_if(timedEvents.begin(), timedEvents.end(), remove_by_caller(widget));
timedEvents.erase(it, timedEvents.end());

注意 lambda 的存在是为了简化这个过程，在 Boost 和 C++11 中都是如此.

Note lambda's exist to simplify this process, both in Boost and C++11.

这篇关于在为每个执行 a 时从 std::vector 中擦除?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！

本站部分内容来源互联网,如果有图片或者内容侵犯了您的权益，请联系我们，我们会在确认后第一时间进行删除！

在为每个执行 a 时从 std::vector 中擦除?

问题描述

推荐答案

相关文档推荐